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Let $F \subset L \subset K$ be fields such that $K/L$ is normal and $L/F$ is purely inseparable. Show that $K/F$ is normal.

Let $\sigma: K \to \overline{K}$ be an $F$-homomorphism. Let $\alpha \in L$ and $p(x) \in F[x]$ such that $p(x) = \min(F,\alpha)$, so $p(\sigma(\alpha))=0$. But $L/F$ is purely inseparable, then $\sigma(\alpha) = \alpha$. Thus, $\sigma$ is a $L$-homomorphism. Since $K/L$ is normal, $\sigma(K) = K$. Therefore, $K/F$ is normal.

I'm just in doubt if I can say that $\alpha$ has a minimal.

First, I tried to get an irreducible polynomial over $F$ that has a root in $K$ and show that it splits on $K$. But I could not show, I wish someone could help me with this idea.

Thanks for the advance.

Lucas
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    I think using splitting field definition is basically rerun the argument you wrote. I think you have to treat $K$ as $F-$vector space. Write out a representation of an element $x\in K$. Then you can just use $Hom_K(K/F,C/F)\to Hom_L(K/L,C/L)\times Hom_F(L/F,C/F)$ is a bijection but $Hom_F(L/F,C/F)={1_F}$ due to purely inseparability. So you can identify the group action on an element. It is not totally clear that $L/K$ normal does implie $F-$homomorphism will fix $K$ but the idea is correct. $a$ has a minimal which is a polynomial in $F[x^p]$ as $L/F$ is purely inseparable. – user45765 May 18 '18 at 13:48
  • However, you know that $a$ is purely inseparable. This says all divisors of minimal polynomial of $p(x)$ must be a factor of the form $(x-a)$ but it must lie in $F[x^p]$(i.e. $(x-a)^{p^m}$ for some $m$. – user45765 May 18 '18 at 13:49
  • Related: https://math.stackexchange.com/q/367344/279515 –  May 31 '19 at 16:15

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Your proof is correct. Since $\alpha \in L$ is algebraic over $F$, it has a minimal polynomial, there is no issue with that.

Notice that you have also shown that if $L/F$ is purely inseparable, then $L/F$ is normal, which is an interesting result. (Compare with What's an example of a non-normal, purely inseparable field extension?.)